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Fair warning: I'm not a mathematician. My graduate education is in physics; I've never had any formal education in abstract algebra, so you may need to explain things like I am five.

Consider the group formed by set $\mathbb Q-\{0\}$ with the binary operation $\times$ (multiplication). This group has an action on the set of real numbers $\mathbb R$. Now consider the equivalence relation defined by that action: $$\forall x,y\in\mathbb R. \quad x\sim y \iff \text{there exists nonzero $q\in\mathbb Q$ such that $x=qy$.}$$

For example, $\frac 56 \sim \frac{11}{17}$, and $2^{1/2} \sim 2^{-1/2}$, but $2^{1/3} \nsim 2^{2/3}$. Some examples of the equivalence classes under this relation are $\{0\}$, $\mathbb Q-\{0\}$, and $\left\{q \cdot \pi \mid q \in \mathbb Q-\{0\}\right\}$.


My question is as follows:

Is it possible to characterize these equivalence classes so that a representative member can be uniquely chosen from each one? More specifically, I would like to construct a function $f:\mathbb R \to \mathbb R$ such that

\begin{align} \forall x\in\mathbb R. &\quad x \sim f(x)\\ \forall x,y\in\mathbb R. &\quad x \sim y \iff f(x)=f(y) \end{align} If such a function cannot be constructed, how can this be proven?

Such a function, if one exists, would allow me to unify an equation that popped up in my research with several others. However, the existence of such a function seems untenable, because in all of those other cases, there were always additional constraints beyond $x\in\mathbb R$ which made picking a representative member straightforward.


I found this existing question but

  1. nearly every word in it flies miles over my head, and
  2. from what I can decipher, it sounds like the question is about addition by rationals. (perhaps all the arguments work out the same; but I cannot tell this due to (1.)!)

Update: I seek a constructive example; something where there is a potentially useful mathematical property which distinguishes the canonical representatives from the rest of the reals. (while a choice function does distinguish them (with regard to fixed points) it isn't particularly useful)

Based on the responses so far I gather that one cannot exist; the trouble then is to prove it.

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  • $\begingroup$ The question you linked is unrelared. Such a function exists if you assume the Axiom of choice (and you could find something weaker than AC that also works). It's kind of like if you look at $\Bbb R$ as a $\Bbb Q$-vector space: you'll have a hard time finding a basis but the Axiom of choice gives you one. $\endgroup$ – xavierm02 Aug 29 '16 at 17:00
  • $\begingroup$ Does the following (implicit) description of such a function $f$ meet your criterion of "construction"? For any $x\in{\mathbb{R}}$, choose a representative $q_{[x]}$ of the equivalence class $[x]$ (this is possible by the axiom of choice) and define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=q_{[x]}$. $\endgroup$ – Janik Aug 29 '16 at 17:02
  • $\begingroup$ Such a function is ensured by the Axiom of Choice, however I think that a constructive example cannot be found. This is because such a function would not be Lebesgue measurable. $\endgroup$ – Crostul Aug 29 '16 at 17:03
  • $\begingroup$ @Crostul I am not familiar with the concept of Lebesgue measurability, can you expand on this? $\endgroup$ – Exp HP Aug 29 '16 at 21:48
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    $\begingroup$ @ExpHP Well, I could explain to you all the details, but I think that these would be too technical for a physicist. I suggest you to look at this , this , and this. In practice, your requested function has similar properties with the Vitali set (second link), which is non-measurable. Now, the existence of non-measurable sets cannot be proved in a constructive way, since there is some model in which every set is measurable and the Axiom of Choice fails. $\endgroup$ – Crostul Aug 29 '16 at 21:58
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The axiom of choice is indeed necessary for this. Specifically:

It is consistent with ZF (= set theory without choice) that there is no transversal of $\sim$ (that is, no set of reals containing exactly one element of each $\sim$-class).

Note that since $\{f(x): x\in\mathbb{R}\}$ would be a transversal of $\sim$, this rules out the possibility of "explicitly" constructing such an $f$.

The proof of course uses forcing and symmetric submodels, so it's a mouthful and I'm not going to give the proof here. But Cohen's original model of $ZF+\neg AC$ is a model in which this is true. See Jech's book on the axiom of choice for details on symmetric submodels, and how this kind of argument goes.

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The function you're trying to create is exactly the kind of thing that the axiom of choice is made for.

Note that there is an (uncountable) collection of (disjoint) equivalence classes whose union is $\Bbb R$. We may denote the collection of these equivalence classes as $\mathcal C = \{S_i : i \in I\}$ (where $I$ is some indexing set). The axiom of choice allows us to say the following:

There exists a function $f: \mathcal C \to \Bbb R$ such that for each $i \in I$, $f(S_i) \in S_i$.

Note that the axiom of choice is somewhat contentious. In particular, there is a consistent, reasonable way of doing mathematics in which we disallow the construction of such sets, and it is impossible to construct (i.e. prove that an example exists) such a function within the confines of the Zermelo-Fraenkel axioms.

I'm not quite sure how one would prove that no explicit example $f$ exists here, but I am fairly confident it can be done. Hopefully this puts you on the right track.


Another point worth considering: it is possible to consider this group via addition. Notably, if we define $$ L_{\Bbb Q} = \{\log(q):q \in \Bbb Q_+\} $$ We can consider the quotient $G = (\mathbb{R},+)/L_{\Bbb Q}$. The map $$ \phi:G \to (\Bbb R, \cdot)/(\Bbb Q \setminus \{0\})\\ \phi([r]) = [\exp(r)] $$ is an injective homomorphism. Here, $[x]$ denotes the equivalence class of $x$.

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  • $\begingroup$ I am a bit lost on the second point; since it is taking a set as an argument, I gather that $\exp$ here is not the typical $\mathbb R \to \mathbb R$ function we all know and love? $\endgroup$ – Exp HP Aug 29 '16 at 17:52
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    $\begingroup$ I think it a bit of an exaggeration nowadays to say that $\mathsf{AC}$ is somewhat contentious. I suspect that it’s not just a minority, but a rather small minority of working mathematicians who have problems with it. $\endgroup$ – Brian M. Scott Aug 29 '16 at 18:31
  • $\begingroup$ @ExpHP: No, $\exp$ is the usual $x\mapsto e^x$, assuming that the logs in question are natural logs. If they’re logs to some other base $b$, then $\exp(r)$ should be interpreted as $b^r$. $\endgroup$ – Brian M. Scott Aug 29 '16 at 18:32
  • $\begingroup$ Yeah, looks like that was poor reading on my part; I mentally swapped $r$ and $[r]$ in the last equation $\endgroup$ – Exp HP Aug 29 '16 at 19:15
  • $\begingroup$ For the record, I did mean natural logs and $e^x$. Thanks Brian for stepping in there. $\endgroup$ – Omnomnomnom Aug 29 '16 at 21:05

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